Problem: Here's a parameterization of a plane: $\vec{v}(x, y) = (4x, -y, 2x + 2y)$ What vectors are normal to the plane $\vec{v}$ ? Choose 2 answers: Choose 2 answers: (Choice A) A $(-4, 2, -8)$ (Choice B) B $(-2, 8, 4)$ (Choice C) C $(2, -8, -4)$ (Choice D) D $(4, -2, 8)$
The vector normal to the area element describes what's perpendicular to the surface, and the vector's magnitude represents the area of a tiny rectangle along the parameterization. We can take multiply the result by $-1$ to find another normal vector. $\text{normal to area element} = \dfrac{\partial \vec{v}}{\partial x} \times \dfrac{\partial \vec{v}}{\partial y}$ Let's calculate the cross product. $\begin{aligned} \dfrac{\partial \vec{v}}{\partial x} \times \dfrac{\partial \vec{v}}{\partial y} &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ 4 & 0 & 2 \\ \\ 0 & -1 & 2 \end{pmatrix} \\ \\ &= 2 \hat{\imath} - 8 \hat{\jmath} - 4 \hat{k} \end{aligned}$ We want two normal vectors. Because the negative of a normal vector is also normal to the surface, we can take the negative of what we just calculated as a second normal vector to the plane $\vec{v}$. Therefore, two vectors normal to $\vec{v}$ are $(2, -8, -4)$ and $(-2, 8, 4)$.